A LevelMathsCambridge (CIE)Pure 3FlashcardsComplex NumbersFurther Complex Numbers

Further Complex Numbers (Cambridge (CIE) A Level Maths: Pure 3): Flashcards

Exam code: 9709

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FrontExponential Form of Complex Numbers
Complete the definition connecting the exponential and trigonometric functions.
$e^{i θ} =______+ i______$

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Cards in this collection (17)

Complete the definition connecting the exponential and trigonometric functions.
$e^{i θ} =______+ i______$
The completed definition is:
$e^{i θ} = \cos θ + i \sin θ$
Read as a complex number, its real part is $\cos θ$ and its imaginary part is $\sin θ$ .
What are the modulus and argument of $e^{i θ}$ ?
Its modulus is $1$ and its argument is $θ$ .
Every value of $e^{i θ}$ therefore lies on the circle of radius $1$ about the origin, at the angle $θ$ .
How do you write a complex number in exponential form?
As $z = r e^{i θ}$ , where $r$ is the modulus and $θ$ is the argument.
The same two quantities that describe a complex number in polar form describe it here, in a shorter notation.
Why is exponential form convenient for multiplying complex numbers?
Because the ordinary index laws apply, so $e^{i θ_{1}} \times e^{i θ_{2}} = e^{i (θ_{1} + θ_{2})}$ .
Multiplying $r_{1} e^{i θ_{1}}$ by $r_{2} e^{i θ_{2}}$ therefore gives $r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$ in a single step.
True or False?
$e^{2 π i}$ and $e^{0}$ are the same number.
True.
Both are equal to $1$ .
Adding $2 π$ to the exponent returns to the same point on an Argand diagram, so $e^{2 k π i} = 1$ for every integer $k$ .
Complete these two standard results.
$e^{i π} =______, e^{\frac{π}{2} i} =______$
The completed results are:
$e^{i π} = - 1, e^{\frac{π}{2} i} = i$
The first is more often written as $e^{i π} + 1 = 0$ , which ties five of the most important constants in mathematics into one equation.
How do you convert $3 e^{\frac{2 π}{3} i}$ into the form $a + b i$ ?
Read off the modulus and the argument, then work out each part separately.
Here the real part is $3 \cos \frac{2 π}{3} = - \frac{3}{2}$ and the imaginary part is $3 \sin \frac{2 π}{3} = \frac{3 \sqrt{3}}{2}$ , giving $- \frac{3}{2} + \frac{3 \sqrt{3}}{2} i$ .
What is the geometric effect of multiplying $z$ by $w = r e^{i θ}$ ?
A stretch from the origin by scale factor $r$ , together with a rotation anticlockwise about the origin through the angle $θ$ .
Both happen at once. A negative $θ$ turns the rotation clockwise instead.
What is the geometric effect of dividing $z$ by $w = r e^{i θ}$ ?
A stretch from the origin by scale factor $\frac{1}{r}$ , together with a rotation clockwise about the origin through the angle $θ$ .
Dividing by $w$ undoes exactly what multiplying by $w$ does.
A point $z$ has modulus $5$ and argument $\frac{π}{4}$ , and is multiplied by $w = 2 e^{\frac{π}{6} i}$ . Complete the result.
$|z w| =______, \arg (z w) =______$
The completed result is:
$|z w| = 10, \arg (z w) = \frac{5 π}{12}$
The two moduli are multiplied and the two arguments added, whichever form the numbers happen to be written in.
True or False?
Dividing $z$ by a complex number always moves it closer to the origin.
False.
Dividing by a number whose modulus is less than $1$ pushes the point further out rather than bringing it in.
Only division by a number with modulus greater than $1$ moves a point closer to the origin.
A complex number is given as $2 \sqrt{3} + 2 i$ . What must you find before describing its geometric effect?
Its modulus and its argument, because a number given in Cartesian form displays neither.
Here $|w| = \sqrt{12 + 4} = 4$ and $\arg w = \tan^{- 1} \frac{2}{2 \sqrt{3}} = \frac{π}{6}$ .
How do you find a square root of $n e^{i α}$ in exponential form?
Square root the modulus and halve the argument.
That gives $\sqrt{n} e^{\frac{α}{2} i}$ , because squaring it returns $n e^{i α}$ by the index laws.
To use the exponential method on $2 + 2 \sqrt{3} i$ , it must be converted first. Complete its exponential form.
$z =______e^{______i}$
The completed form is:
$z = 4 e^{\frac{π}{3} i}$
The modulus is $\sqrt{2^{2} + {(2 \sqrt{3})}^{2}} = 4$ and the argument is $\tan^{- 1} \sqrt{3} = \frac{π}{3}$ .
How do you get the second square root in exponential form?
Add $2 π$ to the argument before halving, which is allowed because it describes the same complex number.
Halving $α + 2 π$ gives $\frac{α}{2} + π$ , so the second root is $\sqrt{n} e^{(\frac{α}{2} + π) i}$ .
True or False?
Converted to Cartesian form, the two roots found by the exponential method are negatives of each other.
True.
Their arguments differ by $π$ , which points them in exactly opposite directions from the origin while leaving the modulus the same.
Written out, they come to $a + b i$ and $- a - b i$ .
Compared with writing the root as $c + d i$ , what does the exponential method avoid?
The pair of simultaneous equations, and the quartic they lead to.
Comparing the moduli and the exponents on the two sides delivers the two unknowns directly, with no equation left to solve.

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